The function f(x) ={[e^(1/x)-1/e^(1/x)+1],x ≠ 0][0,x=0] A. is continuous at x = 0 B. is not continuous at x = 0

Question

The function f(x) = \(\begin{cases} \frac{e^{1/x}-1}{e^{1/x}+1}&, \quad x ≠{0}\\ 0&, \quad \text x =0 \end{cases} \) 

A. is continuous at x = 0 

B. is not continuous at x = 0

C. is not continuous at x = 0, but can be made continuous at x = 0 

D. none of these

Answer 1

Option : (b)

Formula :-

(i) A function f(x) is said to be continuous at a point x = a of its domain, if 

\(\lim\limits_{x \to a}f(x)\) = f(a)

 \(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a)

Given :-

 f(x) = \(\begin{cases} \frac{e^{1/x}-1}{e^{1/x}+1}&, \quad x ≠{0}\\ 0&, \quad x =0 \end{cases} \) 

Using substitution method,

And,

F(0) = 0 

Therefore,

  \(\lim\limits_{x \to 0}f(x)\) ≠ f(0)

Hence, 

f(x) is discontinuous at x = 0